Question

Match List-I with List-II. 

• A. (A)-(II), (B)-(III), (C)-(I), (D)-(IV)
• B. (A)-(II), (B)-(IV), (C)-(I), (D)-(III)
• C. (A)-(III), (B)-(II), (C)-(I), (D)-(IV)
• D. (A)-(I), (B)-(IV), (C)-(II), (D)-(III)

Hint:

Ozonolysis of alkenes results in cleavage of the double bond and formation of carbonyl compounds (aldehydes or ketones), depending on the structure of the alkene.

Answer 1

The Correct Option is B

To address this problem, we must align the isomers of \(C_{10}H_{14}\) from List-I with their corresponding ozonolysis products in List-II. Let's examine each case:

1. (A) Cyclohexene derivative: Ozonolysis of a basic cyclohexene derivative typically yields an aldehyde, resulting from the double bond's cleavage without further substitution complexities.
2. (B) 1,2-Dimethylcyclohexene: This compound's ozonolysis results in a symmetrical cleavage, producing a diketone as both sides of the double bond will bear a carbonyl group.
3. (C) 1-Methylcyclohexene: In this instance, ozonolysis yields both an aldehyde and a ketone. The methyl group's presence facilitates this dual product formation due to differing carbonyl group generation upon cleavage.
4. (D) 1,4-Dimethylcyclohexene: Similar to a basic cyclohexene but with additional methyl groups, this compound primarily forms an aldehyde upon ozonolysis.

Now, let's establish the correct pairings between List-I isomers and List-II ozonolysis products:

• (A) Cyclohexene derivative - (II) Diketone product.
• (B) 1,2-Dimethylcyclohexene - (IV) Aldehyde product.
• (C) 1-Methylcyclohexene - (I) Aldehyde product.
• (D) 1,4-Dimethylcyclohexene - (III) Aldehyde and ketone product.

Therefore, the correct answer is: (A)-(II), (B)-(IV), (C)-(I), (D)-(III).